Current Programmed Control in Power Electronics, Slides of Electrical Engineering

Download Current Programmed Control in Power Electronics and more Slides Electrical Engineering in PDF only on Docsity! Fundamentals of Power Electronics 1 Chapter 12: Current Programmed Control Chapter 12 Current Programmed Control + – Buck converter Current-programmed controller Rvg(t) is(t) + v(t) – iL(t) Q1 L CD1 + – Analog comparator Latch Ts0 S R Q Clock is(t) Rf Measure switch current is(t)Rf Control input ic(t)Rf –+ vref v(t)Compensator Conventional output voltage controller Switch current is(t) Control signal ic(t) m1 t0 dTs Ts on off Transistor status: Clock turns transistor on Comparator turns transistor off The peak transistor current replaces the duty cycle as the converter control input. Docsity.com Fundamentals of Power Electronics 2 Chapter 12: Current Programmed Control Current programmed control vs. duty cycle control Advantages of current programmed control: • Simpler dynamics —inductor pole is moved to high frequency • Simple robust output voltage control, with large phase margin, can be obtained without use of compensator lead networks • It is always necessary to sense the transistor current, to protect against overcurrent failures. We may as well use the information during normal operation, to obtain better control • Transistor failures due to excessive current can be prevented simply by limiting ic(t) • Transformer saturation problems in bridge or push-pull converters can be mitigated A disadvantage: susceptibility to noise Docsity.com Fundamentals of Power Electronics 5 Chapter 12: Current Programmed Control Inductor current waveform, CCM iL(t) ic m1 t0 dTs Ts iL(0) iL(Ts)– m2 buck converter m1 = vg – v L – m2 = – v L boost converter m1 = vg L – m2 = vg – v L buck–boost converter m1 = vg L – m2 = v L Inductor current slopes m1 and –m2 Docsity.com Fundamentals of Power Electronics 6 Chapter 12: Current Programmed Control Steady-state inductor current waveform, CPM iL(t) ic m1 t0 dTs Ts iL(0) iL(Ts)– m2 iL(dTs) = ic = iL(0) + m1dTs d = ic – iL(0) m1Ts iL(Ts) = iL(dTs) – m2d'Ts = iL(0) + m1dTs – m2d'Ts First interval: Solve for d: Second interval: 0 = M 1DTs – M 2D'Ts In steady state: M 2 M 1 = D D' Docsity.com Fundamentals of Power Electronics 7 Chapter 12: Current Programmed Control Perturbed inductor current waveform iL(t) ic m1 t0 DTs Ts IL0 – m2 – m2 m1 Steady-state waveform Perturbed waveform I L0 + iL(0) dTs D + d Ts iL(0) iL(Ts) Docsity.com Fundamentals of Power Electronics 10 Chapter 12: Current Programmed Control Example: unstable operation for D = 0.6 iL(t) ic t0 Ts IL0 iL(0) 2Ts 3Ts 4Ts – 1.5iL(0) 2.25iL(0) – 3.375iL(0) α = – D D' = – 0.6 0.4 = – 1.5 Docsity.com Fundamentals of Power Electronics 11 Chapter 12: Current Programmed Control Example: stable operation for D = 1/3 α = – D D' = – 1/3 2/3 = – 0.5 – 1 8 iL(0) 1 4 iL(0) – 1 2 iL(0) iL(t) ic t0 Ts IL0 iL(0) 2Ts 3Ts 4Ts 1 16 iL(0) Docsity.com Fundamentals of Power Electronics 12 Chapter 12: Current Programmed Control Stabilization via addition of an artificial ramp to the measured switch current waveform + – Buck converter Current-programmed controller Rvg(t) is(t) + v(t) – iL(t) Q1 L CD1 + – Analog comparator Latch ia(t)Rf Ts0 S R Q ma Clock is(t) + + Rf Measure switch current is(t)Rf Control input ic(t)Rf Artificial ramp ia(t) ma t0 Ts 2Ts Now, transistor switches off when ia(dTs) + iL(dTs) = ic or, iL(dTs) = ic – ia(dTs) Docsity.com Fundamentals of Power Electronics 15 Chapter 12: Current Programmed Control Stability analysis: change in perturbation over complete switching periods iL(0) = – dTs m1 + ma iL(Ts) = – dTs ma – m2 iL(Ts) = iL(0) – m2 – ma m1 + ma iL(nTs) = iL((n –1)Ts) – m2 – ma m1 + ma = iL(0) – m2 – ma m1 + ma n = iL(0) αn α = – m2 – mam1 + ma iL(nTs) → 0 when α < 1 ∞ when α > 1 First subinterval: Second subinterval: Net change over one switching period: After n switching periods: Characteristic value: Docsity.com Fundamentals of Power Electronics 16 Chapter 12: Current Programmed Control The characteristic value α • For stability, require | α | < 1 • Buck and buck-boost converters: m2 = – v/L So if v is well-regulated, then m2 is also well-regulated • A common choice: ma = 0.5 m2 This leads to α = –1 at D = 1, and | α | < 1 for 0 ≤ D < 1. The minimum α that leads to stability for all D. • Another common choice: ma = m2 This leads to α = 0 for 0 ≤ D < 1. Deadbeat control, finite settling time α = – 1 – ma m2 D' D + ma m2 Docsity.com Fundamentals of Power Electronics 17 Chapter 12: Current Programmed Control Sensitivity to noise iL(t) ic t0 DTs Ts Steady-state waveform Perturbed waveform dTs (D + d)Ts ic With small ripple: a small amount of noise in the control current ic leads to a large perturbation in the duty cycle. Docsity.com Fundamentals of Power Electronics 20 Chapter 12: Current Programmed Control The first-order approximation iL(t) Ts = ic(t) • Neglects switching ripple and artificial ramp • Yields physical insight and simple first-order model • Accurate when converter operates well into CCM (so that switching ripple is small) and when the magnitude of the artificial ramp is not too large • Resulting small-signal relation: iL(s) ≈ ic(s) Docsity.com Fundamentals of Power Electronics 21 Chapter 12: Current Programmed Control 12.2.1 Simple model via algebraic approach: CCM buck-boost example + – L C R + v(t) – vg(t) Q1 D1 iL(t) iL(t) ic t0 dTs Ts vg L v L Docsity.com Fundamentals of Power Electronics 22 Chapter 12: Current Programmed Control Small-signal equations of CCM buckÐboost, duty cycle control L d iL(t) dt = Dvg(t) + D'v(t) + Vg – V d(t) C dv(t) dt = – D'iL – v(t) R + ILd(t) ig(t) = DiL + ILd(t) Derived in Chapter 7 Docsity.com Fundamentals of Power Electronics 25 Chapter 12: Current Programmed Control The simple approximation, continued Substitute this expression to eliminate the duty cycle from the remaining equations: sCv(s) = – D'ic(s) – v(s) R + IL sLic(s) – Dvg(s) – D'v(s) Vg – V ig(s) = Dic(s) + IL sLic(s) – Dvg(s) – D'v(s) Vg – V Collect terms, simplify using steady-state relationships: sCv(s) = sLD D'R – D' ic(s) – D R + 1 R v(s) – D2 D'R vg(s) ig(s) = sLD D'R + D ic(s) – D R v(s) – D2 D'R vg(s) Docsity.com Fundamentals of Power Electronics 26 Chapter 12: Current Programmed Control Construct equivalent circuit: input port D2 D'R vg + – – D'R D2 D 1 + sL D'R ic D R v ig vg ig(s) = sLD D'R + D ic(s) – D R v(s) – D 2 D'R vg(s) Docsity.com Fundamentals of Power Electronics 27 Chapter 12: Current Programmed Control Construct equivalent circuit: output port sCv(s) = sLD D'R – D' ic(s) – D R + 1 R v(s) – D 2 D'R vg(s) RD' 1 – sLD D'2R ic D R v D2 D'R vg R D sCv vR C Node Docsity.com Fundamentals of Power Electronics 30 Chapter 12: Current Programmed Control Transfer functions predicted by simple model + – ig vg RCr1 f1(s) i c g1 v g2 vg f2(s) i c r2 v + – Gvc(s) = v(s) ic(s) vg = 0 = f2 r2 || R || 1 sC Gvc(s) = – R D' 1 + D 1 – s DL D'2R 1 + s RC 1 + D Control-to-output transfer function Result for buck-boost example Docsity.com Fundamentals of Power Electronics 31 Chapter 12: Current Programmed Control Transfer functions predicted by simple model + – ig vg RCr1 f1(s) i c g1 v g2 vg f2(s) i c r2 v + – Line-to-output transfer function Result for buck-boost example Gvg(s) = v(s) vg(s) i c = 0 = g2 r2 || R || 1 sC Gvg(s) = – D2 1 – D2 1 1 + s RC 1 + D Docsity.com Fundamentals of Power Electronics 32 Chapter 12: Current Programmed Control Transfer functions predicted by simple model + – ig vg RCr1 f1(s) i c g1 v g2 vg f2(s) i c r2 v + – Output impedance Result for buck-boost example Zout(s) = r2 || R || 1 sC Zout(s) = R 1 + D 1 1 + s RC 1 + D Docsity.com Fundamentals of Power Electronics 35 Chapter 12: Current Programmed Control CPM averaged switch model + – L C R + 〈v(t)〉Ts – 〈vg(t)〉Ts 〈iL(t)〉Ts + 〈v2(t)〉Ts – 〈i1(t)〉Ts 〈i2(t)〉Ts Averaged switch network + 〈v1(t)〉Ts – 〈ic(t)〉Ts 〈 p(t)〉Ts Docsity.com Fundamentals of Power Electronics 36 Chapter 12: Current Programmed Control Results for other converters + – L C R + 〈v(t)〉Ts – 〈vg(t)〉Ts 〈iL(t)〉Ts Averaged switch network 〈ic(t)〉Ts 〈 p(t)〉Ts + – L C R + 〈v(t)〉Ts – 〈vg(t)〉Ts 〈iL(t)〉Ts Averaged switch network 〈ic(t)〉Ts 〈 p(t)〉Ts Boost Buck-boost Docsity.com Fundamentals of Power Electronics 37 Chapter 12: Current Programmed Control Perturbation and linearization to construct small-signal model v1(t) Ts = V1 + v1(t) i1(t) Ts = I1 + i1(t) v2(t) Ts = V2 + v2(t) i2(t) Ts = I2 + i2(t) ic(t) Ts = Ic + ic(t) Let V1 + v1(t) I1 + i1(t) = Ic + ic(t) V2 + v2(t) Resulting input port equation: Small-signal result: i1(t) = ic(t) V2 V1 + v2(t) Ic V1 – v1(t) I1 V1 Output port equation: î2 = îc Docsity.com Fundamentals of Power Electronics 40 Chapter 12: Current Programmed Control Expressing the equivalent circuit in terms of the converter input and output voltages + – L C R + – vg ic v– D2 R D R vic D 1 + sL R ig iL i1(s) = D 1 + s L R ic(s) + D R v(s) – D 2 R vg(s) Docsity.com Fundamentals of Power Electronics 41 Chapter 12: Current Programmed Control Predicted transfer functions of the CPM buck converter + – L C R + – vg ic v– D2 R D R vic D 1 + sL R ig iL Gvc(s) = v(s) ic(s) vg = 0 = R || 1 sC Gvg(s) = v(s) vg(s) i c = 0 = 0 Docsity.com Fundamentals of Power Electronics 42 Chapter 12: Current Programmed Control 12.3 A More Accurate Model l The simple models of the previous section yield insight into the low- frequency behavior of CPM converters l Unfortunately, they do not always predict everything that we need to know: Line-to-output transfer function of the buck converter Dynamics at frequencies approaching fs l More accurate model accounts for nonideal operation of current mode controller built-in feedback loop l Converter duty-cycle-controlled model, plus block diagram that accurately models equations of current mode controller l Simulation of current mode control l Comparison of performance: duty-cycle control vs. current-mode control Docsity.com Fundamentals of Power Electronics 45 Chapter 12: Current Programmed Control Linearize IL + iL(t) = Ic + ic(t) – MaTs D + d(t) – M 1 + m1(t) D + d(t) 2 Ts 2 – M 2 + m2(t) D' – d(t) 2 Ts 2 The first-order ac terms are iL(t) = ic(t) – MaTs + DM 1Ts – D'M 2Ts d(t) – D2Ts 2 m1(t) – D'2Ts 2 m2(t) Simplify using dc relationships: iL(t) = ic(t) – MaTsd(t) – D2Ts 2 m1(t) – D'2Ts 2 m2(t) Solve for duty cycle variations: d(t) = 1MaTs ic(t) – iL(t) – D2Ts 2 m1(t) – D'2Ts 2 m2(t) Docsity.com Fundamentals of Power Electronics 46 Chapter 12: Current Programmed Control Equation of the current programmed controller The expression for the duty cycle is of the general form d(t) = Fm ic(t) – iL(t) – Fgvg(t) – Fvv(t) Table 12.2. Current programmed controller gains for basic converters Converter Fg Fv Buck D2Ts 2L 1 – 2D Ts 2L Boost 2D – 1 Ts 2L D' 2Ts 2L Buck-boost D 2Ts 2L – D'2Ts 2L Fm = 1/MaTs Docsity.com Fundamentals of Power Electronics 47 Chapter 12: Current Programmed Control Block diagram of the current programmed controller d(t) = Fm ic(t) – iL(t) – Fgvg(t) – Fvv(t) + – + –– Fm Fg Fv v ic vg d iL • Describes the duty cycle chosen by the CPM controller • Append this block diagram to the duty cycle controlled converter model, to obtain a complete CPM system model Docsity.com Fundamentals of Power Electronics 50 Chapter 12: Current Programmed Control CPM buck-boost converter model + – + –– Fm Fg Fv v ic vg d iL Tv Ti + – I d(t)vg(t) +– L Vg – V d(t) + v(t) – RCI d(t) 1 : D D' : 1 i L(t) Docsity.com Fundamentals of Power Electronics 51 Chapter 12: Current Programmed Control 12.3.2 Solution of the CPM transfer functions v(s) = Gvd(s)d (s) + Gvg(s)vg(s) In the models of the previous slides, the output voltage v can be expressed via superposition as a function of d and vg, through the duty-cycle controlled transfer functions Gvd(s) and Gvg(s): In a similar manner, the inductor current iL can be expressed via superposition as a function of d and vg, by defining the transfer functions Gid(s) and Gig(s): i L(s) = Gid(s)d (s) + Gig(s)vg(s) Gid(s) = i L(s) d (s) vg(s) = 0 Gig(s) = i L(s) vg(s) d(s) = 0 with Docsity.com Fundamentals of Power Electronics 52 Chapter 12: Current Programmed Control System block diagram +–+– – ic Fm CPM controller model d Gvd(s) Gid(s) Gvg(s) Gig(s) + + + + v iL vg Converter transfer functions Fv Fg Docsity.com Fundamentals of Power Electronics 59 Chapter 12: Current Programmed Control 12.3.4 Evaluation of transfer functions Buck converter example • Evaluate general result for the buck converter • Need to evaluate Gvd, Gvg, Gid, Gig Control-to-output transfer function Gvc (s) Line-to-output transfer function Gvg-cpm(s) Ideal current mode control Duty cycle controlGeneral result v i c = R 1 + sRC v vg = 0 Gvd(s) = V D 1 1 + s LR + s 2LC Gvg(s) = D 1 1 + s LR + s 2LC FmGvd 1 + Fm Gid + FvGvd Gvg – FmFgGvd + Fm GvgGid – GigGvd 1 + Fm Gid + FvGvd Docsity.com Fundamentals of Power Electronics 60 Chapter 12: Current Programmed Control Buck converter model Transfer functions Gvd, Gvg, Gid, Gig + – +– L RC 1 : D Id (t) + v(t) – vg(t) i(t) Vgd (t) Gvd(s) = V D 1 den(s) Gvg(s) = D 1 den(s) den(s) = 1 + s LR + s 2LC Gid(s) = V DR 1 + sRC den(s) Gig(s) = D R 1 + sRC den(s) Analyze model to find: All transfer functions of a given circuit have the same poles: Docsity.com Fundamentals of Power Electronics 61 Chapter 12: Current Programmed Control Control-to-output transfer function Gvc(s) Gvc(s) = FmGvd 1 + Fm Gid + FvGvd = Fm V D 1 den(s) 1 + Fm V DR 1 + sRC den(s) + Fv V D 1 den(s) Substitute into expression for Gvc(s): Algebra: Gvc(s) = Fm V D den(s) + FmV DR 1 + sRC + FmFv V D Gvc(s) = Gc0 1 + sQcωc + sωc 2 Docsity.com